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1

Suppose $v \in H^1(T)$ is not essentially constant and $\int_T v \neq 0$. Define $a=-\frac{\int_T v^2}{\int_T v}$. Then $v$ and $a+v$ are orthogonal in $L^2$ (check it) but not in $H^1$ (the derivatives are equal everywhere, so the inner product is the integral of a nonnegative function which is nonzero on a set of positive measure). The first assumption is ...

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Here are some thoughts - not sure if there is a clear-cut answer to your question even for matrices (I use $r(T)$ to denote $\text{rad}(T)$) : $T$ has this property iff $r(T) \geq \|T\|$, which happens iff $$\|T(x)\| \leq r(T)\|x\| \quad\forall x$$ $$\Leftrightarrow \langle Tx,Tx\rangle \leq r(T)^2 \langle x,x\rangle$$ $$\Leftrightarrow r(T)^2I - ... 0 The answer depends on what is meant by "a projection operator". If P is any idempotent operator then, as suggested by user254665, you can write H=\ker(P)\oplus \ker(Id-P). (This follows because for any x\in H you have x-P(x)\in\ker(P).) Hence, P appears to be the "projection operator" onto \ker(Id-P) along \ker(P). On the other hand, if you ... 3 Yes, this does make sense in the context of "rigged Hilbert spaces", e.g., something like what is occasionally called a Gelfand triple H^{+1}\subset L^2\subset H^{-1} of Sobolev spaces on an interval in \mathbb R. Somehow Dirac had a wonderful intuition in this direction already prior to 1930. Also, the possibility of writing "integral kernels" for all ... 1 Here is my understanding. But I know very little about QM, so this explanation may be incorrect. Consider the multiplication operator A defined on L^2[-a, a] by$$A\varphi(x) = x\varphi(x). $$(In quantum mechanics, A corresponds to the position operator.) Since this operator is bounded with the spectrum \sigma(A) = [-a, a], by the spectral ... 0 Since Hilbert spaces are first-countable, it will be enough to check continuity using sequences (as opposed to using nets). We shall base our proof on Lebesgue's dominated convergence theorem (for which using sequences and not nets is essential, as shown here and here). Let x_n \to x in H. Note that the function y \mapsto \Bbb e ^{\Bbb i \langle x_n, y ... 1 OK, I lied :). Then again, I felt like I couldn't really post a question saying "I found a proof scattered over a few Wiki entries and am posting just to collect the info in one place", and that such a statement belonged in an answer, just like this one. I also found this, which proves the result for separable spaces. That proof seems similar, though longer ... 1 Any orthogonal projection is bounded: for any v,$$\|v\|^2=\|Pv+P_\perp v\|^2=\|Pv\|^2+\|P_\perp v\|^2\geq \|Pv\|^2$$(the second equality holding because Pv and P_\perp v are orthogonal), so \|P\|\leq 1. Thus PQ is a composition of bounded operators and hence bounded itself. 3$$ ||PQ||\leq||P|| ||Q||\leq 1.1=1$$Operator norm of orthogonal projection 1 Suppose for now \mu real. Let |\mu|^{\perp} the pushforward of |\mu| with respect to the projection P^{\perp}:H\rightarrow V^{\perp}. Using the hypothesis and a change of variables, we get that for every y\in V^{\perp},$$\widehat{|\mu|^{\perp}}(y) = \int \limits_{V^{\perp}}e^{i(x,y)}d|\mu|^{\perp}(x) = \int \limits_He^{i(P^\perp x,y)}d|\mu|(x) = ...

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The result holds for all Hilbert spaces. The generalization is surprisingly simple. Let us assume, you know the proof for separable Hilbert spaces. Now, let $X$ be any Hilbert space and $\{x_n\} \subset X$ be a bounded sequence. Now, the space $\tilde X = \overline{\operatorname{span}(\{x_n\})}$ is separable and you can invoke the result in this space. ...

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The Dirac formalism is not correct for general selfadjoint operators. It's not true even for the specific operators of Quantum. For example, consider a periodic problem on $[-\pi,\pi]$. The opertor $L=-\frac{d^{2}}{dx^{2}}$ has a two-dimensional eigenspace spanned by $\{ e^{inx},e^{-inx}\}$ for each eigenvalue $\lambda = 1^{2}, 2^{2}, 3^{2},\cdots$ and has ...

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I was working on a similar problem. Here is a way of showing that $C^1[a,b]$ is not a Banach space. We will assume $a=0$ and $b=1$; the method behind the proof below will motivate the general idea. Let $\{f_j\}_{j\geqslant 1}\subset C^1[0,1]$ be a sequence, defined as follows. $$f_j(t)=\left\{ \begin{array}{ll} ... 1 Since A is positive, its spectrum is contained in [0,\infty). Then the spectrum of A+I is contained in [1,\infty); thus 0 is not in the spectrum of A+I and A+I is invertible. To justify that the spectrum translates note that A+I-\lambda I=A-(\lambda-1)I. So A+I-\lambda is not invertible precisely when A-(\lambda -1)I is not invertible, ... 3 It shouldn't be to hard to show that A is injective, for if Ax = 0 then$$\langle x,x \rangle \le \langle Ax,x\rangle =0.$$It will follow that A is invertible once we show that A is surjective: the range R(A) satisfies R(A) = H. Let A^* denote the adjoint of A. Suppose that y \in N(A^*) so that A^*y = 0. Then$$0 = \langle y,A^*y ...

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I take it a pre-Hilbert space is just an inner product space? There are various definitions of "orthonormal basis" that are equivalent in a Hilbert space but not in a pre-Hilbert space; I'm assuming the $e_n$ are orthonormal and have dense span. Then yes, $\langle x,e_n\rangle=0$ for all $n$ does imply $x=0$. Say $\epsilon>0$. Say $y$ is a linear ...

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The adjoint is a map $u^* : Y \rightarrow X$ such that $$\langle ux, y\rangle_Y = \langle x, u^*y\rangle_X$$ for all $x \in X$ and $y \in Y$.

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We need infinite dimensional vector spaces to produce a counterexample. Consider the subspace of $\ell^2$ consisting of sequences $(x_n)$ such that all but finitely many $x_n$ are zero. Take this to be our $U$.

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I had not noticed before that your indices on the matrix were restricted to positive entries. This matrix has the form $$\begin{pmatrix} \alpha_0 & \alpha_{-1} & \alpha_{-2} & \alpha_{-3} & \cdots \\ \alpha_1 & \alpha_{0} & \alpha_{-1} & \alpha_{-2} & \cdots \\ \alpha_2 & \alpha_{1} & ... 3 First remark. QM postulates state that the Hilbert space is separable. Recall that Riesz-Fischer theorem ensures that separable Hilber spaces are all isometrically isomorphic, so it does not really matter which one you choose until you are speaking about the general theory. In concrete realizations, it is obviously useful pick the Hilbert that best reflects ... 1 For each finite subset F \subset A, define$$ s_F := \sum_{a\in F} |\varphi(a)|^2 $$Since \alpha:= \sum_{a\in A} |\varphi(a)|^2 < \infty, for any \epsilon > 0, \exists F_0 \subset A finite such that$$ |s_F - \alpha| < \epsilon^2 \quad\forall F\supset F_0 \text{ finite} $$Now define \psi \in D by$$ \psi(a) = \varphi(a) \text{ if } a\in ...

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Since that is a convergent series, all but a countable number of $\phi(a)$ are $0$, also any arrangement of those ones $a_1, \dots, a_n, \dots$ has limit $0$. So taken a function $\phi$, $\forall \varepsilon >0$ you can always find a $\psi \in D$ such that $d(\phi,\psi) < \varepsilon$: you just need to take $\psi(a)=0$ if $a \neq a_n$ for all $n$ or ...

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$D$ is a vector space that contains the functions $\phi_a$ defined by $\phi_a(a)=1, \phi_a(x)=0$ for $x\neq a$. The closed linear span of the latter set is $l^2$. So the closure of $D$ contains $l^2$ (and $l^2$ contains $D$). (assuming $A$ is countable.. maybe it still works anyway)

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It is exactly the polarization identity, $(x,y) = x\bar y$ is the standard Hermitian inner product on $\mathbb{C}$. Note that $x\bar x = |x|^2$ if $x \in \mathbb{C}$. Alternatively, just expand everything on the left hand side (using $x\bar x = |x|^2$) and do lots of tedious algebra, essentially recreating the proof of polarization.

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The problem is that $Y_1$ and $Y_2$ may have two different norms so $Y_1$ need not to be isomorphic to $Y_2$. Consider for example $c_{00}$-the space of all sequences $(a_n)$ such that $a_k=0$ for large $k$ and $Y_1=\ell^1$, $Y_2=\ell^2$ (the space of all summable/square summable sequences). Let $f_1,f_2$ be the natural inclusions: then all your conditions ...

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Since the Gram matrix defining the inner product (it has as entry at place $(i,j)$ the inner product of $e_i$ by $e_j$) is symmetric and positive definite (and conversely), you can characterize it with Sylvester's criterion: all principal minors should be positive. In the case of a symmetric $2\times 2$ matrix, say $$\begin{bmatrix} a & b \\ b & c ... 7 By definition an inner product \langle\cdot,\cdot\rangle on a real vector space V is a bilinear, symmetric and positive definite form. In the case of V=\mathbb{R}^n all inner products have form \langle x,y\rangle=x^TAy, where A is a symmetric n\times n matrix with n positive eigenvalues. 0 There are two important aspects to a Hilbert space: the inner product and the completeness property. I don't have much of a background in analysis so I can't explain very much about completeness, but I can tell you why the inner product is important for the Fourier transform. The inner product is important because it gives us the concept of orthogonality, ... 0 If A: H\to H is a continuous linear operator and  g :\mathbb R\to\mathbb R is a discontinuous function at t_0\in\mathbb R, then the family U(t)=g(t)A does not satisfy the condition:$$\|g(t)A\varphi-g(t_0)A\varphi\|=|g(t)-g(t_0)|\cdot \|A\varphi\|$$does not converge to 0. 1 If A is a subspace of an inner product space X, then there are two kinds of projections onto A: orthogonal projection \mathcal{O} and closest point projection \mathcal{C}. One type of projection exists iff the other does and, in that case, the two must be equal. Furthermore, the projection is always unique if it exists. That part is exactly as it ... 0 question is good formulated A \ne B , i got close values when i was working something so i tought it was equal, but its not, it was close. 0 Example...... contrary to the claim in a comment. f(t) = e^{-t^2/2}. Then$$ \int_{-\infty}^\infty |f(t)|^2\,dt = \int_{-\infty}^\infty e^{-t^2}\,dt = \sqrt{\pi} \approx 1.772454 $$but$$ \sum_{n=-\infty}^\infty |f(n)|^2 = \sum_{n=-\infty}^\infty e^{-n^2} =\vartheta_3(0,e^{-1}) \approx 1.772637 $$Not equal. 1 To begin with, we have to correctly interpret what f(n) means in this context. For example, if f\in L^2(dx), we can change f on the integers and not change the L^2 norm of f at all. For example, let g(x)=f(x) if x is not an integer and let g(n)=0 then ||f||_{L^2}=||g||_{L^2} but the \ell^2 norms of the associated sequences are certainly ... 1 l^2 are precisely square integrable sequences if you endow set of all integers (or natural numbers) with counting measure. In this way l^2 can be seen as special case of L^2, while summation (both finite and that of absolutely convergent series) is special case of integration. And in case you don't know: counting measure assigns to every element of ... 1 Let T = p-q. Note that p(H) and p(H)^\perp are invariant subspaces of T. Consider the restriction of T to p(H). We can write$$ T\mid_{p(H)} = \operatorname{id} - q $$It should be easy to show that this operator is positive (I'll leave it to you). Of course, the restriction of T to p(H)^\perp is 0, which is a positive operator. Note ... 1 By definition, we have$$ F^{\perp} = \{y \in X: \forall x \in F, (x|y) = 0\}\\ F^{\perp \perp} = \{y \in X: \forall x \in F^\perp, (x|y) = 0 \} $$Now, if x \in F, then by definition of F^\perp: (x|y)= (y|x) = 0 for all y in F^\perp. So, if x \in F, then it must be that x \in F^{\perp \perp}. 0 Meanwhile I got it.. :) Equality On the one hand:$$\varphi\in\big(\bigcup_{\lambda\in\Lambda}A_\lambda\big)^\perp\iff\varphi\perp\bigcup_{\lambda\in\Lambda}A_\lambda\iff\varphi\perp A_\lambda\quad(\forall\lambda\in\Lambda)\varphi\perp A_\lambda\quad(\forall\lambda\in\Lambda)\iff\varphi\in ...

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You are over thinking it. Assume $z_0\in K^\perp$ have the said properties. Let $d = \|x - y_0\| = \rho(x, K) = \langle x,z_0 \rangle$. Notice that $$x = (x - d z_0) + d z_0,$$ which is obviously the orthogonal decomposition of $x$: $$\langle x - d z_0, d z_0 \rangle = d\langle x, z_0 \rangle - d^2 \|z_0\|^2 = d^2 - d^2 = 0.$$ That is, for every $y\in ... 2 First thing to note is that, eigen values of$\hat{n}\cdot \vec{\sigma}$is 1 and -1. Because, $$\hat{n}\cdot \vec{\sigma} = \begin{bmatrix} n_3 & n_1-n_2i\\ n_1 + n_2i & -n_3 \end{bmatrix}$$ $$\det(\hat{n}\cdot \vec{\sigma}-\lambda I) = \lambda^2 - (n_1^2 + n_2^2 + n_3^2) = \lambda^2 - 1 = 0$$ By spectral decomposition theorem, we have two ... 1 First, I am not sure what the author means by Banach's theorem, but it seems like the relevant result from functional analysis which he's using is the bounded inverse theorem: if$T:(X,\left\|\right\|_{X})\rightarrow (Y,\left\|\right\|_{Y})$is a continuous (i.e. bounded) bijective linear map between two Banach spaces$X$and$Y$, then$T^{-1}: ...

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Often, the best way to establish weak convergence is a two step procedure: 1) Show that the sequence in question is bounded. In this case, this is straightforward (do it!). 2) Show that there is a dense subset $X \subset L^2$ for which $\langle x_n, y\rangle \to \langle x, y\rangle$ holds for all $y \in X$, where $x$ is the (supposed) weak limit. This ...

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After looking in my Griffiths quantum mechanics book, I think I now realize that the Hermitian Operator defines what a Hermitian Matrix is. Anyways, I wonder if the following properties is enough to define what a Hermitian Matrix is. 1) All eigenvalues are real 2) The eigenvectors are mutually orthogonal (Inner product between any of them yields $0$). 3) ...

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By definition, a complex square matrix $A$ is hermitian if and only if $A^H = A$. Where $A^H = \bar A^T$ denotes the conjugate transposed of $A$. It is the result of the spectral theorem, that all eigenvalues of a hermitian matrix are real and there exists eigenvector basis. The matrix $$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$ has clearly ...

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Proof sketch / Hints: Since the equation is linear, $\langle f, g \rangle = \langle f, \operatorname{id} \rangle$ holds for every $f\in V$ if and only if it holds every $f$ in the orthonormal basis you gave above. Thus, we are looking at $$\langle e_i, g \rangle = \langle e_i, \operatorname{id} \rangle$$ for $i = 1,\dotsc, 5$. Now, using $g = ... 0 I think I found my answer, please correct it if I am wrong. In many Physics books, they mention the "completeness relation", $$\int \partial E \ \ | E \rangle \langle E | = \mathbb{1}$$ where$ \mathbb{1} $is the identity operator. So I think I need to prove this in order to demonstrate completeness. I have worked with wavelets, so below lead to my ... 2 This is an answer for$\mathsf{Hilb}_1$. Let$X$be any nonzero Hilbert space. I claim the coproduct$X \vee X$doesn't exist in$\mathsf{Hilb}_1$. Assume that it did, call it$X \xrightarrow{i_0} X \vee X \xleftarrow{i_1} X$. Then by the universal property of the coproduct, we get an isometry$f : X \vee X \xrightarrow{(\operatorname{id}, ...

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